# One Way ANOVA Table for N = 18, k = 3 & α = 5%

One way ANOVA table or summary to estimate the test of hypothesis (H_{0}) at 5% or 0.05 significance level for sample *means* by using *F-distribution table*

One-Way ANOVA Classification Table | ||||
---|---|---|---|---|

Sources of Variation | df | SS | MSS | F-Ratio |

between Treatments | 2 | 10.4111 | 10.4111/2 = 5.2056 | 5.2056/3.2467 = 1.6034 |

Error | 15 | 48.7 | 48.7/15 = 3.2467 | |

Total | 17 |

## Work with Steps for N = 18, k = 3 & α = 5%

The below is the example work with steps shows how to generate the one way ANOVA table for the different means of treatments to find the F test statistic to test hypothesis (H

step 1 Address the formula input parameters and values

x

x

x

step 2 Form the below table to carry out the analysis of

Individual Squares :

step 3 Find Correction Factor (CF)

N = total number of elements in all three sample dataset.

N = 7 + 5 + 6

N = 18

Correction factor (CF) = G

CF = 212

= 44944/18

CF = 2496.8889

step 4 Find the sum of squares of all individual items (∑x

Total sum of squares (TSS) = ∑∑x

= 2556 - 2496.8889

TSS = 59.1111

step 5 Find the sum of squares between the treatments (SST)

SST = k ∑ i = 1 T

= (∑x

= 7056/7+2809/5+5625/6 - 2496.8889

= (1008 + 561.8 + 937.5) - 2496.8889

= 2507.3 - 2496.8889

= 10.4111

step 6 Find sum of squares due to the error (SSE)

SSE = TSS - SST

= 59.1111 - 10.4111

= 48.7

step 7 The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table.

One-Way ANOVA Classification Table

step 8 Find Critical value of F

The critical value of F

For treatments, the critical value F

There is no significance difference between treatments, since the calculated value of F

_{0}) at a 5% or 0.05 significance level for three sample means x_{1}, x_{2}& x_{3}.

__Workout :__

step 1 Address the formula input parameters and values

x

_{1}= 10, 12, 13, 11, 10, 13 & 15

x

_{2}= 9, 11, 13, 9 & 11

x

_{3}= 11, 15, 13, 12, 14 & 10

step 2 Form the below table to carry out the analysis of

*variance*

Total | square | ||||||||

x_{1} | 10 | 12 | 13 | 11 | 10 | 13 | 15 | ∑x_{1} = 84 | (∑x_{1})^{2} = 7056 |

x_{2} | 9 | 11 | 13 | 9 | 11 | ∑x_{2} = 53 | (∑x_{2})^{2} = 2809 | ||

x_{3} | 11 | 15 | 13 | 12 | 14 | 10 | ∑x_{3} = 75 | (∑x_{3})^{2} = 5625 | |

G = 212 | ∑T_{i}^{2} = 15490 |

Individual Squares :

Total | ||||||||

x_{1}^{2} | 100 | 144 | 169 | 121 | 100 | 169 | 225 | ∑x_{1}^{2} = 1028 |

x_{2}^{2} | 81 | 121 | 169 | 81 | 121 | ∑x_{2}^{2} = 573 | ||

x_{3}^{2} | 121 | 225 | 169 | 144 | 196 | 100 | ∑x_{3}^{2} = 955 | |

∑x_{ij}^{2} = 2556 |

N = total number of elements in all three sample dataset.

N = 7 + 5 + 6

N = 18

Correction factor (CF) = G

^{2}N

CF = 212

^{2}18

= 44944/18

CF = 2496.8889

step 4 Find the sum of squares of all individual items (∑x

_{ij}) and then total sum of squares (TSS)

Total sum of squares (TSS) = ∑∑x

_{ij}

^{2}- CF

= 2556 - 2496.8889

TSS = 59.1111

step 5 Find the sum of squares between the treatments (SST)

SST = k ∑ i = 1 T

_{i}

^{2}n

_{i}- CF

= (∑x

_{1})

^{2}n

_{1}+(∑x

_{2})

^{2}n

_{2}+(∑x

_{3})

^{2}n

_{3}- CF

= 7056/7+2809/5+5625/6 - 2496.8889

= (1008 + 561.8 + 937.5) - 2496.8889

= 2507.3 - 2496.8889

= 10.4111

step 6 Find sum of squares due to the error (SSE)

SSE = TSS - SST

= 59.1111 - 10.4111

= 48.7

step 7 The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table.

Sources of Variation | df | SS | MSS | F-ratio |

between Treatments | k - 1 | SST | SST/k-1 = MST | MST/MSE = F _{T} |

Error | N - k | SSE | SSE/N-k = MSE | |

Total | N - 1 |

One-Way ANOVA Classification Table

Sources of Variation | df | SS | MSS | F-Ratio |
---|---|---|---|---|

between Treatment | 3-1 = 2 | 10.4111 | 10.4111/2 = 5.2056 | 5.2056/3.2467 = 1.6034 |

Error | 18-3 = 15 | 48.7 | 48.7/15 = 3.2467 | |

Total | 18-1 = 17 |

step 8 Find Critical value of F

_{e}or Table value of F

The critical value of F

_{e}from the F-distribution table for the degrees of freedom (k - 1, N - k) at 5% significance level.

For treatments, the critical value F

_{e}(2,15) from F-distribution table at 5% of significance level is 3.6823

__Inference__There is no significance difference between treatments, since the calculated value of F

_{0}= 1.6034 is smaller than the table value of F

_{e}= 3.6823. Therefore the null hypothesis H

_{0}is accepted.

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